A Level · Mathematics 9709 · Pure Mathematics 3

Complex numbers: A Level Mathematics questions with answers

Free Complex numbers questions for A Level Mathematics (9709), part of Pure Mathematics 3. Each question has the answer and the marking points an examiner would look for.

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Complex numbers practice questions

Question 1Structured · 13 marksA* challenge

The region R in an Argand diagram consists of points z satisfying |z − 3 − 4i| ≤ 3 and |z| ≤ |z − 6i|.

(a)

Sketch R on an Argand diagram.

[2]
(b)

Find the least value of arg z for points in R, giving the exact value of tan(arg z), and the point at which it occurs.

[4]
(c)

Find the greatest value of arg z for points in R.

[2]
(d)

Find the greatest value of |z| for points in R.

[2]
(e)

Find the area of R.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)disc centre 3 + 4i radius 3 (touching the imaginary axis), below the line Im z = 3
circle centre (3, 4), radius 3, and line Im z = 3 B1
correct region shaded B1
2
(b)tan = 724 (16.3°); z = 9625 + 2825i
tangent from O: arg z = α − β with tan α = 43, sin β = 35 M1
tan(α − β) = 724, 16.3° (0.284 rad) A1
|z| = 4 at the point of contact M1
z = 9625 + 2825i (lies below Im z = 3, so in R) A1
4
(c)86.7° (1.51 rad), where z = 3 − 2√2 + 3i
meets Im z = 3 where (x − 3)2 = 8, x = 3 − 2√2 M1
tan−1(33 − 2√2) = 86.7° (1.51 rad) A1
2
(d)√(26 + 12√2) = 6.56
furthest point is the corner 3 + 2√2 + 3i (the far side of the circle lies above Im z = 3) M1
√(26 + 12√2) = 6.56 A1
2
(e)9 cos−1(13) − 2√2 ≈ 8.25
R is a segment; half-angle at the centre φ with cos φ = 13 M1
area = ½ × 9 × 2φ − ½ × 9 sin 2φ = 9φ − 2√2 M1
8.25 A1
3
Question 2Structured · 5 marks

The complex numbers u and w are given by u = 4 + 5i and w = 3 − i.

(a)

Find uw in the form x + iy.

[2]
(b)

Find the modulus and argument of u. Give the argument in radians correct to 3 significant figures.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)17 + 11i
Expand with i2 = −1 M1
Correct answer A1
2
(b)|u| = 6.4, arg u = 0.896
Modulus B1
Argument method in correct quadrant M1
0.896 A1
3
Question 3Structured · 13 marksA* challenge

The roots of the equation z2 − (4 + 2i)z + c = 0, where c is a complex constant, are represented by the points P and Q in an Argand diagram, and the sum of the roots by the point S. O is the origin.

(a)

Given that OPSQ is a square, find c.

[4]
(b)

Given instead that OPSQ is a rhombus with angle POQ = 2θ, find c in terms of θ.

[4]
(c)

Describe the locus of the point representing c as the rhombus varies.

[2]
(d)

For a rhombus OPSQ as in (b), find the value of c for which one of the roots is real, and find both roots in this case.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)6 + 8i
p + q = 4 + 2i and the diagonals of the square bisect each other at right angles M1
p, q = (2 + i) ± i(2 + i) M1
p = 1 + 3i, q = 3 − i A1
c = pq = 6 + 8i A1
4
(b)c = (3 + 4i)sec2 θ
diagonals perpendicular: p, q = (2 + i) ± ki(2 + i) M1
k = tan θ A1
pq = (2 + i)2(1 + tan2 θ) M1
c = (3 + 4i)sec2 θ A1
4
(c)half-line arg c = tan−1(43), |c| > 5
part of the half-line from O with arg c = tan−1(43) (53.1°) B1
|c| > 5 (|c| ≥ 5 if the degenerate case P = Q is included) B1
2
(d)c = 154 + 5i; roots 52 and 32 + 2i
Im((2 + i)(1 + ik)) = 0 gives k = −½ M1
roots 52 and 32 + 2i A1
c = 154 + 5i A1
3
Question 4Structured · 6 marksA* challenge

The complex numbers u and w are given by u = 4 − i and w = −3 + i.

(a)

Find uw in the form x + iy, where x and y are exact.

[3]
(b)

Find the modulus and argument of uw, giving the modulus correct to 3 significant figures and the argument in radians correct to 3 decimal places in the range −π < θ ≤ π.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)−1310 − 110i
multiply by conjugate M1
denominator 10 A1
correct x and y A1
3
(b)1.3, −3.065
|u| ÷ |w| M1
1.3 A1
−3.065 A1
3
Question 5Structured · 11 marksA* challenge
(a)

Find the two square roots of the complex number 32i, giving your answers in the form x + iy, where x and y are real.

[5]
(b)

The complex number z satisfies |z − (2 + 2i)| = 1. Find the greatest and least possible values of |z|, giving your answers in exact form.

[3]
(c)

The equation z3 − 7z2 + 16z − 10 = 0 has a root 3 + i. Find the other two roots.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)±(4 + 4i)
(x + iy)2 = 32i, equate real and imaginary parts M1
x2 − y2 = 0 and 2xy = 32 A1
eliminate one variable to a quadratic in x2 or y2 M1
solve M1
±(4 + 4i) A1
5
(b)√8 + 1, √8 − 1 (3.83, 1.83)
|2 + 2i| = 2.83 B1
add and subtract the radius M1
3.83 and 1.83 A1
3
(c)3 − i and 1
conjugate 3 − i is a root B1
quadratic factor z2 − 6z + 10 M1
real root 1 A1
3
Question 6Structured · 5 marks

The complex numbers u and w are given by u = 5 − 4i and w = −3 − 3i.

(a)

Find uw in the form x + iy.

[2]
(b)

Find the modulus and argument of u. Give the argument in radians correct to 3 significant figures.

[3]
Show answers and mark scheme
PartAnswer and marking pointsMarks
(a)−27 − 3i
Expand with i2 = −1 M1
Correct answer A1
2
(b)|u| = 6.4, arg u = −0.675
Modulus B1
Argument method in correct quadrant M1
−0.675 A1
3

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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in